Question 270821: how many liters of at 10% solution must be mixed with a 50% acid solution in order to obtain 30 liters of a 30% solution? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! how many liters of at 10% solution must be mixed with a 50% acid solution in order to obtain 30 liters of a 30% solution?
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One variable solution:
Equation:
active + active = active
0.10x + 0.50(30-x) = 0.30*30
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Multiply thru by 100 to get:
10x + 50*30 - 50x = 30*30
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-40x = -20*30
x = 15 liters (amt. of 10% solution needed in the mixture)
50-15 = 35 liters (amt. of 50% solution needed in the mixture)
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Cheers,
Stan H.
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